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Transcript
Electric Forces, Fields, Potential and Energy Fundamental Charge: The charge on one electron. e = 1.6 x 10 -19 C Unit of charge is a Coulomb (C) Two types of charge: Positive Charge: A shortage of electrons. Negative Charge: An excess of electrons. Conservation of charge – The net charge of a closed system remains constant. Nucleus - - n + n + + n + n n + + n - - - - Negative Neutral Atom Atom Positive Atom Number Numberof ofelectrons electrons><=Number Numberof ofprotons protons Number of electrons Number of protons -19 -2e = -3.2 x 10 CC +2e = +3.2 x 10-19 Electric Forces Like Charges - Repel F + + Unlike Charges - Attract - F F + F Coulomb’s Law – Gives the electric force between two point charges. q1q2 F k 2 r Inverse Square Law k = Coulomb’s Constant = 9.0x109 Nm2/C2 q1 = charge on mass 1 q2 = charge on mass 2 r = the distance between the two charges The electric force is much stronger than the gravitational force. Example 1 Two charges are separated by a distance r and have a force F on each other. qq F k F 1 2 2 r q2 q1 F r If r is doubled then F is : ¼ of F If q1 is doubled then F is : 2F If q1 and q2 are doubled and r is halved then F is : 16F Example 2 Two 40 gram masses each with a charge of 3μC are placed 50cm apart. Compare the gravitational force between the two masses to the electric force between the two masses. (Ignore the force of the earth on the two masses) 3μC 40g 3μC 40g 50cm m1m2 Fg G 2 r 6.67 10 11 (.04)(.04) 2 (0.5) 4.27 10 q1q2 FE k 2 r 6 6 ( 3 10 )( 3 10 ) 9 9.0 10 (0.5) 2 13 N 0.324 N The electric force is much greater than the gravitational force Example 3 Three charged objects are placed as shown. Find the net force on the object with the charge of -4μC. F k - 5μC 45º 20cm 202 202 28cm q1q2 r2 (5 106 )(4 106 ) F1 9 10 4.5N 2 (0.20) 9 (5 106 )(4 106 ) F2 9 10 2.30 N 2 (0.28) 9 F1 45º - 4μC 5μC 20cm F2 F1 and F2 must be added together as vectors. F1 2.3cos45≈1.6 45º F2 2.3sin45≈1.6 F1 = < - 4.5 , 0.0 > + F2 = < 1.6 , - 1.6 > Fnet = < - 2.9 , - 1.6 > - 1.6 - 2.9 29º θ 3.31 Fnet 2.9 2 1.6 2 3.31N 1.6 tan 29 2.9 1 3.31N at 209º Example 4 Two 8 gram, equally charged balls are suspended on earth as shown in the diagram below. Find the charge on each ball. 20º L = 30cm FE q 10º 10º L = 30cm 30sin10º q r r =2(30sin10º)=10.4cm 2 q1q2 q FE k 2 k 2 r r FE Draw a force diagram for one charge and treat as an equilibrium problem. T FE q T sin 80 .08 .08 T .081N sin 80 Tsin80º 80º Tcos80º Fg = .08N FE T cos 80 q2 k (. 081 ) cos 80 .104 2 .014 2 q (.104) 2 k q 1.3 10 7 C Electric Fields: A property of space around a charge that causes forces on other charges. The force per unit charge felt by a positive test charge placed in the field + FE Use a positive test charge FE _ FE + + FE FE + + FE + + Use a positive test charge The gravitational Force = (The mass)xx(The (TheElectric Gravitational The Electric Force = (The charge) Field) Field) FgE mg qE q1eqm2 M FFg EGk 22 rr M e FEFg q Eg kG 2 r r2 q m The units forfor a gravitational field N/kg The units a electric field areare N/C Things to Know • Electric field lines go out of a positive charge and into a negative charge. • Positive charges experience a force with the field. • Negative charges experience a force against the field. • The Electric Field inside a conductor is zero. Example 3 What is the electric field at a point 25cm away from a point charge of 2µC? q Ek 2 r 6 2 10 E 9 10 (.25) 2 9 5 N E 2.88 10 C Away from the charge 2µC 25cm E Example 4 A 5 gram mass with a charge of -3µC is released from rest in a constant field of 20000N/C directed in the positive x direction. Find its position and velocity after 4.0 seconds. Negative charges experience a force against the field FE = qE F ma FE ma qE ma qE m a 12 2 m s 5g -3µC m v v0 at (12)( 4) 48 s 1 2 x x0 v0t at 2 1 x (12)( 4) 2 96m 2 Example 6 What is the electric field at the origin? E3 EE21 kq kq1 1 EE32 1 222 rr 666 10 10 (9(910 109 9)()(121..5010 ) )) EE32 1 222 5 . 0 34.5 NN 720 EE32 1844 735 CC To the right (+) Up (+) To the right (+) The electric field due to each charge must be calculated individually and then added together as vectors. Now add the vectors together: E3=844N/C E1 = < 735, 0 > E2 = < 720, 0 > E3 = < E E1=735N/C θ 1455N/C E2=720N/C 844N/C 0 , 844 > E = < 1455, 844 > E 14552 844 2 N E 1682 C 1 844 tan 1455 30 The electric field due to more than one point charge. F + F Two charged parallel plates The Electric field is constant between the two plates. The Electric Field inside a conductor is zero E = 0N/C An electron has a velocity of 6x105m/s as it enters a constant electric field of 10000N/C as shown in the diagram below. How far will the electron travel in the x direction before striking one of the plates? 2.82 x 10-3m _ v 4cm FNow First mawe have projectile motion 1a simple 2 acceleration the force and of the1 2 y y0 electron v0t must at x x v t at problem! 0 0 be calculated. FE ma 2 2 5 shown 9 ma The qE Electron path as 1 will follow15a parabolic 2 x (6 10 )( 4.7 10 ) .02 (1.8 10 19 )t qE2 (1.6 10 )(10000) 315 m x 2.82 a 1 .810 10 m 2 31 s t 4m .7 10 9 s9.1110 Let’s make a parallel comparison of Gravitational Potential Energy (Ug) to Electric Potential Energy (UE). U g mgh Units: U E qEd Units: N kg m Nm J kg N C m Nm J C The Electric Potential Energy at a point is the work done to move a charge from infinity to that point. Now we will do an example: An electron is released from rest in an electric field of 2000N/C. How fast will the electron be moving after traveling 30cm? E0 W E f v=? _ v = 0m/s _ 30cm UE K 1 2 qEd mv 2 2qEd v m 2(1.6 1019 )( 2000)(.3) v 9.111031 m v 1.45 10 s 7 A new quantity is defined called the Electric Potential Difference. It is the work done per unit charge to move a small positive test charge between two points. Electric Potential Difference UE V q qEd V q V Ed Electric Potential Energy Ch arg e The Electric Potential Difference can also called the Electric Potential, the Potential, or the voltage. Units: J volt V C Remember Energy and voltage are scalars, so you don’t have to deal with vectors (direction) The electric potential energy can now be written in terms of the electric potential. U E qEd V Ed U E qV U E qV Let’s do an example using this new concept: The potential difference between two charge plates is 500V. Find the velocity of a proton if it is accelerated from rest from one plate to the other. E W E 0 High Low Potential + Potential - + - + + - + - + - 500V Positive charges move from high to low potential Negative charges move from low to high potential f UE K 1 2 qV mv 2 2qV v m 2(1.6 10 19 )(500) v 1.67 10 27 5 m v 3.110 s Example: E0 W E f U E K1 K 2 q1q2 1 2 1 2 k mv mv r 2 2 q2 k mv 2 r 2 kq v2 rm Two 40 gram masses each with a charge of -6µC are 20cm apart. If the two charges are released, how fast will they be moving when they are a very, very long way apart. (infinity) kq2 (9 109 )(6 10 6 ) 2 m v 6.36 rm (.2)(. 04) s Let’s use the electric potential energy between two charges to derive an equation for the electric potential (voltage) due to a single point charge. The electric potential energy is given by: q2P r q1 q1q2 UE k r Let’s remove charge q2 and consider the electric potential (voltage) at point P. kq1q2 UE q1 r V k q q2 r To find the potential due to more than one point charge simply add up all the individual potentials: i q V k r i Example 1: The electron in the Bohr model of the atom can exist at only certain orbits. The smallest has a radius of .0529nm, and the next level has a radius of .212nm. a) What is the potential difference between the two levels? b) Which level has a higher potential? q V k r e V1 k r1 r1 r2 19 1.6 10 V1 (9 10 ) 27.2V 9 .0529 10 19 1 . 6 10 V2 (9 109 ) 6.79V 9 .212 10 +e 9 r1 is at a higher potential. potential diff V 27.2 6.79 20.4V Example 2 What is the electric potential at the center of the square? 45º 45º r r r 2k V V qq2 kr rr 0.10 6 6 10 5 2 r 9 10 10 9.01 10 V V 9 910 071 . . 071 m r .071 65 J 10 3410 V V 16..27 C 2 r r V k i 2 qi r Vtotal 1.27 106 (1.27 106 ) 6.34 105 6.34 105 J Vtotal 1.27 10 C 6 Example 3: A proton is moved from the negative plate to the positive plate of a parallel-plate arrangement. The plates are 1.5cm apart, and the electric field is uniform with a magnitude of 1500N/C. a) How much work would be required to move a proton from the negative to the positive plate? b) What is the potential difference between the plates? c) If the proton is released from rest at the positive plate, what speed will it have just before it hits the negative plate? W U E qEd N W (1.6 10 C )(1500 )(.015m) C W 3.6 10 18 J 19 Example 3: A proton is moved from the negative plate to the positive plate of a parallel-plate arrangement. The plates are 1.5cm apart, and the electric field is uniform with a magnitude of 1500N/C. b) What is the potential difference between the plates? V Ed N V (1500 )(.015m) C J V 22.5 C Example 3: A proton is moved from the negative plate to the positive plate of a parallel-plate arrangement. The plates are 1.5cm apart, and the electric field is uniform with a magnitude of 1500N/C. c) If the proton is released from rest at the positive plate, what speed will it have just before it hits the negative plate? Use conservation of energy E0 W E f UE K 1 2 qV mv 2 2qV v m 2(1.6 10 19 )( 22.5) v 1.67 10 27 m v 6.57 10 4 s Example 4: Compute the energy necessary to bring together the charges in the configuration shown below: Calculate the electric potential energy between each pair of charges and add them together. qq12q1qq332 UU13 2312kk rr 66 6 666 ( 4 10 )( 4 10 ( 4 10 )( 4 10 )) 99 9 UU13 (910 10 )) 2312(9 .2 UU13 72JJ 231200..72 U total .72 J (.72 J ) (.72 J ) U total .72 J